A point P moves such that the distance from ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the point $P$ be $(x, y)$. Given $Q(0,2)$ and $R(-1,0)$,the condition is $PQ = \frac{1}{\sqrt{2}} PR$. Squaring both sides,we get $2(PQ)^2 = (

Vedclass · Accepted Answer

a circle with centre at $(1,4)$ and radius $\sqrt{10}$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$. Given $Q(0,2)$ and $R(-1,0)$,the condition is $PQ = \frac{1}{\sqrt{2}} PR$. Squaring both sides,we get $2(PQ)^2 = (PR)^2$. Substituting the coordinates,$2(x^2 + (y-2)^2) = (x+1)^2 + y^2$. Expanding the terms: $2(x^2 + y^2 - 4y + 4) = x^2 + 2x + 1 + y^2$. $2x^2 + 2y^2 - 8y + 8 = x^2 + 2x + 1 + y^2$. Rearranging the terms: $x^2 + y^2 - 2x - 8y + 7 = 0$. This is the equation of a circle. The centre is $(-\frac{-2}{2}, -\frac{-8}{2}) = (1, 4)$. The radius is $\sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 4^2 - 7} = \sqrt{1 + 16 - 7} = \sqrt{10}$. Thus,the locus is a circle with centre $(1, 4)$ and radius $\sqrt{10}$.

$A$ point $P$ moves such that the distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is

Similar Questions

If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2x-4y-20=0$ and $x^2+y^2-4x+2y-44=0$ is $2:3$,then the locus of $P$ is a circle with centre :

The locus of the centre of a circle which passes through the origin and cuts off a length of $4$ units from the line $x=3$ is

The locus of the point whose ratio of distance from the origin to its distance from $(-2, -3)$ is $5: 7$,is given by:

The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is

If a point $P(x, y)$ moves such that the sum of the squares of its coordinates is equal to their product,then the locus of $P$ excluding the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module